# Cannot determine the P-value [closed]

I have a chi squared of 0.1185 with 1 degree of freedom. How can i determine the P-value?

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Because $\chi^2(1)$ is the distribution of the square of a standard normal variate, your question is tantamount to asking for the probability that a standard normal value exceeds $\sqrt{0.1185}\approx0.34$ in size. That's an elementary calculation (and I'm sure you have software or accurate tables to carry it out). –  whuber Mar 5 '12 at 23:32

## closed as not a real question by whuber♦Aug 14 '12 at 12:38

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By p-value I presume you mean "the probability of seeing a value of my test statistic as large as this, if it really comes from a $\chi^2$ distribution with one degree of freedom". So if you get a low p-value it means it is an improbable value to have seen.

In R:

1-pchisq(0.1185, 1)

or

pchisq(0.1185, 1, lower=FALSE)

In Excel

=1-CHISQ.DIST(0.1185,1, TRUE)

or

=CHISQ.DIST.RT(A2,1)

The p-value is very large because your value of the test statistic is actually considerably less than the average value of such a $\chi^2$ random variable. If anything, it is unusually small. So basically, you do not have a large value and hence there is not evidence to dismiss the hypothesis that it genuinely comes from a $\chi^2$ distribution.

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so, what is the p-value? –  user176105 Mar 5 '12 at 22:54
I would recommend you always use FALSE instead of F, so it should be pchisq(0.1185, 1, lower = FALSE). –  Henrik Mar 5 '12 at 23:08
@user176105 I'll leave it up to you to calculate... you must have access to either R or Excel, surely. –  Peter Ellis Mar 5 '12 at 23:10
@PeterEllis so my teacher just gave me a table like this: sociology.ohio-state.edu/people/ptv/publications/p%20values/… so the only thing i can do is narrow my p-value to a range? –  user176105 Mar 5 '12 at 23:14
Linear interpolation of the 2-tailed p-value for $z=\sqrt{0.1185}$ gives $0.73$ which is fully accurate to two d.p. (In R this can be done as (sqrt(0.1185)-.39)/(.25-.39) * (.80-.70) + 0.70; the numbers 0.39, 0.25, 0.80, and 0.70 come from your teacher's table.) Alternatively, applying Simpson's Rule (x<-0.1185;1 - sqrt(x)*(2*exp(-x/2) + 4)/(3*sqrt(2*pi))) gives 0.7306, which is within 0.0001 of the truth: no table is needed at all. –  whuber Mar 5 '12 at 23:43
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The upper tail of the $\chi^2_1$ distribution, beyond 0.1185 is

pchisq(0.1185, df=1, lower=F)
[1] 0.7306671